# Lorentz Transformation via Geometry

Today, I was tutoring and explained the space-time. I explained how one can convert North-South into West-East by rotating, and how you can convert time into space with velocity.

Below the Energy-Momentum stuff the book had some problems. One was the following:

Given an event at (1 Lightsecond, 2 s), give it's coordinates in a frame of reference that is moving with 0.6 of speed of light to it.

This the diagram I came up with, in blue there is the event. (Not really to scale.)

http://wstaw.org/m/2011/11/11/m8.png

Then I projected the point onto the orange axes, and I get smaller values on each. This makes sense to me, as a moving observer would see thing smaller and slower, resulting in less values in either measurement.

But how do I get the orange lines? I cannot see how I get this angle. If I got them, I could tell you the speed of the reference frame, but the speed is given here …

And did I do the projection of the speed right?

• It's so rare that a scanned sketch is as visually appealing as this. – Alan Rominger Nov 11 '11 at 18:43
• It is not scanned in the first place :-). I made it with Xournal and a Wacom Intous. – Martin Ueding Nov 16 '11 at 7:38

I'm going to take $c=1$. The top orange line, which is the $t'$ axis in your drawing, is drawn with a slope of $1/0.6$; this follows from the definition of velocity. The bottom orange line, which is the $x'$ axis, has a slope of 0.6.

What's a little harder is to get the scale on the orange axes. The scale follows from the fact that area is preserved by Lorentz transformations. The velocity of 0.6 turns out to correspond to a Lorentz transformation in which a square is distorted into a parallelogram with its long axis stretched by a factor of 2, and its short axis contracted by $1/2$.

If you want to see this approached developed in more detail, see ch. 7 of this online book. (I'm the author.)

There are a couple of recent commercial textbooks that use similar geometric approaches:

• Mermin, It's About Time: Understanding Einstein's Relativity

• Takeuchi, An Illustrated Guide to Relativity

• I think I found the formula you refer to in my book: $\hat{x'}=\frac{(v/c, 1)}{\gamma}$ So the lines has the slope of $v/c$ and is streched by $\gamma$. But that would be 0.8, not 0.5. Is that the wrong formula? – Martin Ueding Nov 12 '11 at 14:28

the $ct'$ and $x'$ axes are orthogonals in the Minkowski sense: $(ct')_{\mu} (x')^{\mu} = 0$. So Far So good, then to measure the length of the point of intersection in $ct'$ you need to draw a parabola passing at that point (not any parabola, the constant length parabola). The intersection at your $ct$ axis gives you the proper length of the time coordinate in the primed system.

In other words $(ct')_{\mu} (ct')^{\mu} = (ct'_\text{proper})^2$

Likewise for the $x'$ coordinate.